python cannot call module method from staticmethod?

Question

consider the following pattern:

'''some module body'''

def __foo():
    '''module method foo'''
    pass

class Dummy(object):
    @staticmethod
    def bar():
        __foo()

__foo() # No Error.
Dummy.bar() #NameError: Global name "_Dummy__foo" is not defined.

Why is this happening?

--

And if it's bad to name it with "__", what is the best practice in Python to make module methods available only for inner-module functions/methods?

cf. the Python documentation. "Any identifier of the form __spam (at least two leading underscores, at most one trailing underscore) is now textually replaced with _classname__spam, where classname is the current class name with leading underscore(s) stripped." — jason
– jason, Commented Jul 5, 2013 at 3:43
Better reference to the docs (not for an obsolete version): docs.python.org/2/tutorial/… — Chris Morgan
– Chris Morgan, Commented Jul 5, 2013 at 3:44
The "best practice in Python to make module methods available only to inner-module functions/methods" is don't do it at all. — BrenBarn
– BrenBarn, Commented Jul 5, 2013 at 3:57
It is nearly impossible to make anything truly private in Python. This is by design. Instead, put a single underscore in front of anything clients of the module shouldn't touch, and don't touch anything from other modules that starts with an underscore. — user2357112
– user2357112, Commented Jul 5, 2013 at 4:00

user2357112 · Accepted Answer · 2013-07-05 03:41:20Z

5

Don't start names with double underscores. Any identifier found in a class statement, starting with at least 2 underscores and ending with less than 2 underscores, gets _Classname prepended to it, where Classname is the name of the class. This is supposed to provide limited support for private-ish variables, but using it is often considered bad practice.

answered Jul 5, 2013 at 3:41

user2357112

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1 Comment

Ignacio Vazquez-Abrams Over a year ago

Technically it's for MI collision prevention (not that MI is such a great idea regardless...).

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